Calculus I. George Voutsadakis 1. LSSU Math 151. Lake Superior State University. 1 Mathematics and Computer Science

Transcription

1 Calculus I George Voutsadakis 1 1 Mathematics and Computer Science Lake Superior State University LSSU Math 151 George Voutsadakis (LSSU) Calculus I November / 77

2 Outline 1 Applications of the Derivative Linear Approximations Extreme Values The Mean Value Theorem and Monotonicity The Shape of a Graph L Hôpital s Rule Graph Sketching and Asymptotes Applied Optimization Newton s Method Antiderivatives George Voutsadakis (LSSU) Calculus I November / 77

3 Linear Approximations Subsection 1 Linear Approximations George Voutsadakis (LSSU) Calculus I November / 77

4 Linear Approximations Linear Approximation of f If f is differentiable at x=a and x is small, then the difference y=f(a+ x) f(a) can be approximated by y f (a) x. George Voutsadakis (LSSU) Calculus I November / 77

5 Linear Approximations Examples 1 Use a Linear Approximation to approximate ; Consider f(x)= 1 x ; Note that f (x)= 1 ; We would like to x 2 approximate f(10+0.2) f(10); According to the linear approximation scheme, f(10.2) f(10) f (10) x= = Approximate the value of 3 8.1; Consider f(x)= 3 x; Note that f (x)= 1 3 x 2/3 ; We would like to approximate f(8.1); According to the linear approximation scheme, f(8.1) f(8) f (8) x= /3 0.1= Therefore = George Voutsadakis (LSSU) Calculus I November / 77

6 Linear Approximations Applied Example: Thermal Expansion A thin metal cable has length L=12 cm when the temperature is T=21 C. Estimate the change in length when T rises to 24 C, assuming that dl dt =kl, where k= C 1 (k is the coefficient of thermal expansion); Consider the function L(T); Note that dl dt =kl; We would like to approximate L(24) L(21); According to the linear approximation scheme, dl L(24) L(21) dt T=21 T=kL(21) T= = = George Voutsadakis (LSSU) Calculus I November / 77

7 Linear Approximations Applied Example: Pizza Gimmicks A pizza company at Corleone claims that its pizzas are circular with diameter 50 cm. Find the area of each pizza and estimate the quantity lost or gained if the diameter is off by at most 1.2 cm; Consider the function A(d)=π( d 2 )2 = π 4 d2 ; Note that A (d)= π 2 d; Thus, A(50)= π =625π cm 2. Next, We would like to approximate A, which we take to be A(51.2) A(50) or A(50) A(48.8); According to the linear approximation scheme, A A (d) d= π 2 d d=π =30π cm2. George Voutsadakis (LSSU) Calculus I November / 77

8 Linear Approximations Linearization of a Function If f is differentiable at x=a and x is a value close to a, then f(x) may be approximated by f(x) L(x)=f (a)(x a)+f(a). The straight line L(x) is called the linearization of f(x) at x=a; Example: Compute the linearization of f(x)= xe x 1 at a=1; We have f(1)= 1e 0 = 1; Moreover, f (x)= ex 1 2 x + xe x 1, whence f (1)= = 3 2 ; Therefore, L(x) = f (1)(x 1)+f(1) 3 = 2 (x 1)+1= 3 2 x 1 2. George Voutsadakis (LSSU) Calculus I November / 77

9 Linear Approximations Error Estimates If y=f(x) is a function, then the percentage error (in decimal) incurred by estimating f(a+ x) by using the linearization of f is the absolute value of the actual error divided by the actual value: PErr= f(a+ x) f(a) f (a) x. f(a+ x) Example: Estimate tan( π ) and compute the percentage error of the estimation; Let f(x)=tanx; Thus, f (x)=sec 2 x; For a= π 4 and x=0.02, we have PErr = f(a+ x) f(a) f (a) x f(a+ x) = tan(π ) tan π 4 sec2 π tan( π ) = George Voutsadakis (LSSU) Calculus I November / 77

10 Extreme Values Subsection 2 Extreme Values George Voutsadakis (LSSU) Calculus I November / 77

11 Extreme Values Absolute Extrema on an Interval Let f(x) be a function on an interval I and a I; We say that f(a) is the absolute minimum of f(x) on I if f(a) f(x) for all x I; absolute maximum of f(x) on I if f(x) f(a) for all x I; The process of finding the max or min values, collectively referred to as extreme values, is called optimization; Absolute Extrema on a Closed Interval A continuous function f(x) on a closed interval I=[a,b] attains both a minimum and a maximum value on I. George Voutsadakis (LSSU) Calculus I November / 77

12 Extreme Values Local Extrema We say that f(x) has a local (or relative) minimum at x=c if f(c) is the minimum value of f on some open interval containing c; local (or relative) maximum at x=c if f(c) is the maximum value of f on some open interval containing c; Observe that at the local extrema, if the derivative (slope of the tangent line) exists, it has value 0. George Voutsadakis (LSSU) Calculus I November / 77

13 Extreme Values Critical Points of a Function f(x) A number c in the domain of f(x) is a critical point if f (c)=0 or f (c) does not exist; Example: Consider f(x)=x 3 9x 2 +24x 10; We have f (x)=3x 2 18x+24=3(x 2 6x+8)=3(x 4)(x 2). Therefore, the critical points of f(x) are x=2 and x=4; Example: Consider f(x)= x ; We saw that f 1, if x<0 (x)={ Therefore, the only critical 1, if x>0 point of f(x) is x=0; Fermat s Theorem for Local Extrema If f(c) is a local min or max, then c is a critical point of f(x). Caution: The converse is not true!! Even though c might be a critical point, f(c) might not be a local extremum. George Voutsadakis (LSSU) Calculus I November / 77

14 Extreme Values Optimization in a Closed Interval Extreme Values on Closed Interval If f is continuous on[a,b] and f(c) is an extreme value of f on[a,b], then either c is a critical point or one of the endpoints a or b. Example: Find the extrema of f(x)=2x 3 15x 2 +24x+7 on[0,6]; We have f (x)=6x 2 30x+24= 6(x 2 5x+4)=6(x 4)(x 1). Therefore, the critical points of f in [0,6] are x=1 and x=4; Now, we compute: f(0)=7,f(1)= 18,f(4)= 9,f(6)=43; Thus, in [0,6], f(4) is the absolute min and f(6) is the absolute max of f; George Voutsadakis (LSSU) Calculus I November / 77

15 Extreme Values Another Example Find the max value of f(x)=1 (x 1) 2/3 on[ 1,2]; We have f (x)= 2 3 (x 1) 1/3 = x 1 ; Thus, in[ 1,2], f has the critical point x=1; We have f( 1)=1 3 4, f(1)=1 and f(2)=0; Therefore, f(1) is the absolute max and f( 1) the absolute min of f in[ 1,2]; How about the extreme values of f(x)=x 2 8lnx on[1,4]? George Voutsadakis (LSSU) Calculus I November / 77

16 Extreme Values A Third Example Find the max value of f(x)=sinx+cos 2 x on[0,2π]; We have f (x)= cosx 2cosxsinx= cosx(1 2sinx); Thus, in[0,2π], f has the critical points x= π 2, 3π 2 and x= π 6, 5π 6 ; We have f(0)=1, f( π 6 )= 5 4, f(π 2 )=1, f(5π 6 )= 5 4, f( 3π 2 )= 1, f(2π)=1; Therefore, f( 3π 2 ) is the absolute min and f( π 6 )=f(5π 6 ) the absolute max of f in[0,2π]; George Voutsadakis (LSSU) Calculus I November / 77

17 Extreme Values Rolle s Theorem Rolle s Theorem If f(x) is continuous on[a,b] and differentiable on(a,b), and f(a)=f(b), then there exists a c (a,b), such that f (c)=0. George Voutsadakis (LSSU) Calculus I November / 77

18 Extreme Values Illustration of Rolle s Theorem Verify Rolle s Theorem for f(x)=x 4 x 2 on[ 2,2]; Clearly, f(x) is continuous in[ 2,2], since it is a polynomial function; It is differentiable on( 2, 2), with derivative f (x)=4x 3 2x=2x(2x 2 1); Therefore, by Rolle s Theorem, there exists a c between 2 and 2, such that f (c)=2c(2c 2 1)=0. Actually, there are three solutions c=0, or c=± 2 2 ; George Voutsadakis (LSSU) Calculus I November / 77

19 Extreme Values Using Rolle s Theorem to Prove Uniqueness of Roots Show that the equation x 3 +9x 4 has precisely one real root. Consider the function f(x)=x 3 +9x 4; Since f(0)= 4<0, whereas f(1)=6>0, by the Intermediate Value Theorem, there exists a c in (0,1), such that f(c)=c 3 +9c 4=0; Therefore, the given equation has at least one real solution; Assume that it has two real solutions a and b, with a<b; This means that f(a)=f(b)=0; Since f is continuous on[a,b] and differentiable on(a,b), with derivative f (x)=3x 2 +9, by Rolle s Theorem, there exists c (a,b), such that f (c)=3c 2 +9= 0; But this is impossible!! Therefore, f(x)=0 has at most one real solution, i.e., it has exactly one real solution! George Voutsadakis (LSSU) Calculus I November / 77

20 The Mean Value Theorem and Monotonicity Subsection 3 The Mean Value Theorem and Monotonicity George Voutsadakis (LSSU) Calculus I November / 77

21 The Mean Value Theorem and Monotonicity The Mean Value Theorem The Mean Value Theorem If f(x) is continuous on the closed interval[a,b] and differentiable on the open interval(a,b), then there exists at least one value c in(a,b), such that f (c)= f(b) f(a) ; b a George Voutsadakis (LSSU) Calculus I November / 77

22 The Mean Value Theorem and Monotonicity An Example Consider f(x)= x; This function is continuous on[1, 9] as a root function; Moreover, it is differentiable on(1, 9), with derivative f (x)= 1 2 ; By the x MVT, we may conclude that there exists at least one c (1,9), such that f (c)= f(9) f(1) ; This equation says that 2 c = 1 4 ; Solving this, yields c=4; George Voutsadakis (LSSU) Calculus I November / 77

23 The Mean Value Theorem and Monotonicity Sign of the Derivative and Monotonicity A function f(x) is increasing on(a,b) if f(x 1 )<f(x 2 ), for all x 1 <x 2 in(a,b); decreasing on(a,b) if f(x 1 )>f(x 2 ), for all x 1 <x 2 in(a,b); f(x) is monotonic on(a,b) if it is either increasing or decreasing on (a,b); Sign of the Derivative If f is differentiable on(a,b), then: If f (x)>0 for all x (a,b), then f is increasing on(a,b); If f (x)<0 for all x (a,b), then f is decreasing on(a,b); Geometrically, the theorem says that the sign of the slopes of the tangent lines to y=f(x) determines the kind of the monotonicity of f on(a,b); George Voutsadakis (LSSU) Calculus I November / 77

24 The Mean Value Theorem and Monotonicity First Derivative Test for Critical Points First Derivative Test If f(x) is differentiable and that c is a critical point of f(x), then: if f (x) changes from+to at c, then f(c) is a local maximum; if f (x) changes from to+at c, then f(c) is a local minimum; Example: Let f(x)=x 2 2x 3; we have f (x)=2x 2=2(x 1); Thus, x= 1 is a critical point of f; The following is a sign table for f (x) and summarizes the behavior of f(x) with respect to monotonicity; Function x<1 x>1 f (x) + f(x) It shows that f(1) is a local minimum; George Voutsadakis (LSSU) Calculus I November / 77

25 The Mean Value Theorem and Monotonicity Second Example Consider the function f(x)=x 3 27x 20; We have f (x)=3x 2 27=3(x 2 9)=3(x+3)(x 3); Therefore, f(x) has critical points at x=±3; The following is a sign table for f (x) and summarizes the monotonicity of f(x): Function x< 3 3<x<3 x>3 f (x) + + f(x) It shows that f( 3) is a local maximum and f(3) is a local minimum; George Voutsadakis (LSSU) Calculus I November / 77

26 The Mean Value Theorem and Monotonicity Third Example Consider the function f(x)=cos 2 x+sinx in(0,π); We have f (x)= 2cosxsinx+cosx= cosx(1 2sinx); Therefore, f(x) has critical points at x= π 2 and x=π 6, 5π 6. The following is a sign table for f (x) and summarizes the monotonicity of f(x): Function 0<x< π π 6 6 <x<π π 2 2 <x<5π 5π 6 6 <x<π f (x) + + f(x) It shows that f( π 6 ),f(5π 6 ) are local maxima and f(π 2 ) is a local minimum; George Voutsadakis (LSSU) Calculus I November / 77

27 The Shape of a Graph Subsection 4 The Shape of a Graph George Voutsadakis (LSSU) Calculus I November / 77

28 The Shape of a Graph Concavity and Inflection Points If f(x) is differentiable in an open interval(a,b), then f is concave up on(a,b) if f (x) is increasing on(a,b); f is concave down on(a,b) if f (x) is decreasing on(a,b); Test for Concavity If f (x) exists for all x (a,b), then: If f (x)>0 for all x (a,b), then f is concave up on(a,b); If f (x)<0 for all x (a,b), then f is concave down on(a,b); The points where concavity changes are called inflection points; Test for Inflection Points Assuming that f (c) exists, if f (c)=0 and f (x) changes sign at x=c, then f(x) has an inflection point at x=c. George Voutsadakis (LSSU) Calculus I November / 77

29 The Shape of a Graph First Example Study the function f(x)=cosx on[0,2π] with respect to monotonicity and concavity; We have f (x)= sinx; Therefore, f(x) has critical points at x=0,π,2π; We have f (x)= cosx; Therefore, f (x) zeros at x= π 2, 3π 2 ; The following is a combined sign table for f (x),f (x) and summarizes the monotonicity and concavity of f(x): Function 0<x< π π 2 2 <x<π π<x<3π 3π 2 2 <x<2π f (x) + + f (x) + + f(x) It shows that f(0), f(2π) are local maxima, f(π) is a local minimum and at π 2, 3π 2 f has inflection points; George Voutsadakis (LSSU) Calculus I November / 77

30 The Shape of a Graph Second Example Study the function f(x)=3x 5 5x 4 +1 with respect to monotonicity and concavity; We have f (x)=15x 4 20x 3 =5x 3 (3x 4); Therefore, f(x) has critical points at x=0, 4 3 ; We have f (x)=60x 3 60x 2 =60x 2 (x 1); Therefore, f (x) zeros at x=0,1; The following is a combined sign table for f (x),f (x) and summarizes the monotonicity and concavity of f(x): 4 3 <x Function x<0 0<x<1 1<x< 4 3 f (x) + + f (x) + + f(x) It shows that f(0) is a local maximum, f( 4 3 ) is a local minimum and at x = 1 f has an inflection point; George Voutsadakis (LSSU) Calculus I November / 77

31 The Shape of a Graph Second Derivative Test for Critical Points Second Derivative Test Let c be a critical point of f(x). If f (c) exists, then: if f (c)>0, then f(c) is a local minimum; if f (c)<0, then f(c) is a local maximum; if f (c)=0, the test is inconclusive. George Voutsadakis (LSSU) Calculus I November / 77

32 The Shape of a Graph Example of Second Derivative Test Use Second derivative test to analyze the critical points of f(x)=(2x x 2 )e x ; f (x)=(2 2x)e x +(2x x 2 )e x = (2 x 2 )e x ; Therefore x=± 2 are the critical points; Moreover, f (x)= 2xe x +(2 x 2 )e x = (2 2x x 2 )e x ; Thus, f ( 2)>0 and f ( 2)<0, showing that f has a local min f( 2) and a local max f( 2); George Voutsadakis (LSSU) Calculus I November / 77

33 The Shape of a Graph One More Example Use Second derivative test to analyze the critical points of f(x)=x 5 5x 4 ; f (x)=5x 4 20x 3 =5x 3 (x 4); Therefore x=0,4 are the critical points; Moreover, f (x)=20x 3 60x 2 =20x 2 (x 3); Thus, f (0)=0 and f (4)>0, showing that f has a local min f(4) but the Second Derivative Test for x=0 is inconclusive; One needs to revert to the signs of the first derivative! George Voutsadakis (LSSU) Calculus I November / 77

34 L Hôpital s Rule Subsection 5 L Hôpital s Rule George Voutsadakis (LSSU) Calculus I November / 77

35 L Hôpital s Rule L Hôpital s Rule Assume that f(x),g(x) are differentiable on an open interval containing a, that f(a)=g(a)=0 and that g (x) 0 (except perhaps at a); Then f(x) lim x ag(x) = lim f (x) x ag (x), assuming that the limit on the right exists or is infinite; The same conclusion holds if f(x),g(x) are differentiable for x near, but not equal to, a and limf(x)=± and lim g(x)=± ; x a x a The rule may also be applied for one-side limits; George Voutsadakis (LSSU) Calculus I November / 77

36 L Hôpital s Rule Applying L Hôpital s Rule I x 3 8 Compute lim x 2 x 4 +2x 20 ; Set f(x)=x 3 8 and g(x)=x 4 +2x 8; We have f(2)=0=g(2); Therefore, by L Hôpital s Rule, lim x 2 f(x) g(x) = lim x 2 f (x) g (x) = lim x 2 3x 2 = = x 3 +2 George Voutsadakis (LSSU) Calculus I November / 77

37 L Hôpital s Rule Applying L Hôpital s Rule II 4 x 2 Compute lim x 2 sin(πx) ; Set f(x)=4 x 2 and g(x)=sin(πx); We have f(2)=0=g(2); Therefore, by L Hôpital s Rule, lim x 2 f(x) g(x) = lim x 2 f (x) g (x) = lim x 2 2x πcos(πx) = 4 π. George Voutsadakis (LSSU) Calculus I November / 77

38 L Hôpital s Rule Applying L Hôpital s Rule III cos 2 x Compute lim x π1 sinx ; 2 Set f(x)=cos 2 x and g(x)=1 sinx; We have f( π 2 )=0=g(π 2 ); Therefore, by L Hôpital s Rule, lim x π 2 f(x) g(x) = lim x π 2 = lim x π 2 f (x) g (x) 2cosxsinx cosx = lim x π 2 (2sinx) = 2. George Voutsadakis (LSSU) Calculus I November / 77

39 L Hôpital s Rule The Form 0 Compute lim x 0 + (xlnx); Note that lim x 0 +x=0 and lim x 0 +lnx= ; The form 0 is indeterminate; The following method may be used to lift this indeterminacy: Rewrite xlnx= lnx 1 x ; Set f(x)=lnx and g(x)= 1 x ; We have lim x 0 +lnx= and lim x x =+ ; Therefore, by L Hôpital s Rule, lim x 0 + f(x) g(x) = lim x 0 + f (x) g (x) = lim x x 1 x 2 = lim x 0 +( x)=0. George Voutsadakis (LSSU) Calculus I November / 77

40 L Hôpital s Rule Using Rule Twice Evaluate lim x 0 e x x 1 cosx 1 ; lim x 0 e x x 1 cosx 1 = ( 0 0 ) = lim x 0 (e x x 1) (cosx 1) = lim x 0 e x 1 sinx = ( 0 0 ) = lim x 0 (e x 1) ( sinx) = lim x 0 e x cosx = 1. George Voutsadakis (LSSU) Calculus I November / 77

41 L Hôpital s Rule Care Needed Before Applying L Hôpital s Rule Evaluate lim x 1 x x+1 ; x lim 2 +1 x 12x+1 = 2 3. We do not want to apply L Hôpital s Rule, when the form we are dealing with is not 0 0 or ; Careless application in the example above would yield the INCORRECT CONCLUSION x lim x = lim x 12x+1 x 1 2 = limx=1 2 x 1 3. Error!! George Voutsadakis (LSSU) Calculus I November / 77

42 L Hôpital s Rule The Form Compute lim( 1 x 0 sinx 1 x ); 1 Note that lim x 0 sinx = and lim x 0 1 x = ; The form is indeterminate; The following method may be used to lift this indeterminacy: 1 Rewrite sinx 1 x = x sinx xsinx ; Set f(x)=x sinx and g(x)=xsinx; We have lim x 0 (x sinx)=0 and lim x 0 xsinx=0; Therefore, by L Hôpital s Rule, lim x 0 f(x) g(x) = lim x 0 f (x) g (x) = lim x 0 1 cosx sinx+xcosx =(0 0 ) = lim x 0 sinx cosx+cosx xsinx = 0 2 =0. George Voutsadakis (LSSU) Calculus I November / 77

43 L Hôpital s Rule The Form 0 0 Compute lim x 0 + xx ; Consider f(x)=x x ; Then Therefore lim x 0 +lnf(x) = lim x 0 +ln(x x ) = lim x 0 +(xlnx) = lim x 0 + lnx 1 = lim x 0 + x 1 x 1 x 2 = lim x 0 +( x)=0. lim x 0 +f(x)= lim x 0 +elnf(x) =e lim x 0 + lnf(x) =e 0 =1. George Voutsadakis (LSSU) Calculus I November / 77

44 L Hôpital s Rule L Hôpital s Rule for Limits at Infinity Suppose that f(x),g(x) are differentiable in an interval(b, ) and that g (x) 0 for x>b; If lim f(x) and lim g(x) exist and either both are zero or both x x infinite, then f(x) lim x g(x) = lim f (x) x g (x) assuming that the limit on the right exists; An analogous result holds for x ; George Voutsadakis (LSSU) Calculus I November / 77

45 L Hôpital s Rule Form Suppose we want to discover which of the two functions f(x)=x 2 and g(x)=xlnx grows faster as x ; To this end, we compute x lim 2 x xlnx ; x lim 2 x x xlnx = lim x lnx =( )= lim x 1 1 x = lim x x= ; Thus f(x) grows asymptotically faster than g(x); Suppose we want to discover which of the two functions f(x)=(lnx) 2 and g(x)= x grows faster as x ; To this end, we compute lim x ; x (lnx) 2 lim x x (lnx) 2 = ( )= lim x = ( )= lim x 1 2 x 2lnx x 1 2 x 4 x Thus g(x) grows asymptotically faster than f(x); = lim x x 4lnx = lim x x 8 = ; George Voutsadakis (LSSU) Calculus I November / 77

46 L Hôpital s Rule Super-polynomial Growth of e x We show that f(x)=e x grows asymptotically faster than any power function f(x)=x n, for fixed n; We calculate lim x e x x n = lim x e x nx n 1 = lim x e x n(n 1)x n 2 = lim x e x n(n 1)(n 2)x n 3 = = lim x e x n! = 1 n! lim x e x =. Therefore f(x) grows asymptotically faster than g(x); George Voutsadakis (LSSU) Calculus I November / 77

47 Graph Sketching and Asymptotes Subsection 6 Graph Sketching and Asymptotes George Voutsadakis (LSSU) Calculus I November / 77

48 Graph Sketching and Asymptotes Graph Sketching I Study the function f(x)=x 2 4x+3 and sketch its graph; We have f (x)=2x 4=2(x 2); Therefore, f(x) has a critical point at x=2; We have f (x)=2; Therefore, f (x) does not have any zeros; The following is a combined sign table for f (x),f (x) and summarizes the monotonicity and concavity of f(x): Function x<2 2<x f (x) + f (x) + + f(x) It shows that f(2)= 1 is a local minimum; George Voutsadakis (LSSU) Calculus I November / 77

49 Graph Sketching and Asymptotes Graph Sketching II Study the function f(x)= 1 3 x3 1 2 x2 2x+3 and sketch its graph; We have f (x)=x 2 x 2=(x+1)(x 2); Therefore, f(x) has critical points at x= 1,2; We have f (x)=2x 1; Therefore, f (x) has a zero x= 1 2 ; The following is a combined sign table for f (x),f (x) and summarizes the monotonicity and concavity of f(x): Function x< 1 1<x< <x<2 2<x f (x) + + f (x) + + f(x) It shows that f(2) is a local minimum, f( 1) is a localmaximumandatx= 1 2 f hasaninflectionpoint; 1 George Voutsadakis (LSSU) Calculus I November / 77

50 Graph Sketching and Asymptotes Graph Sketching III Study the function f(x)=3x 4 8x 3 +6x 2 +1 and sketch its graph; We have f (x)=12x 3 24x 2 +12x=12x(x 1) 2 ; Therefore, f(x) has critical points at x=0,1; We have f (x)=36x 2 48x+12= 12(3x 2 4x+1)=12(x 1)(3x 1); Therefore, f (x) has zeros x= 1 3,1; The following is a combined sign table for f (x),f (x) and summarizes the monotonicity and concavity of f(x): Function x<0 0<x< <x<1 1<x f (x) f (x) f(x) It shows that f(0) is a local minimum, and at x= 1 3,1 f has inflection points; 1 George Voutsadakis (LSSU) Calculus I November / 77

51 Graph Sketching and Asymptotes Graph Sketching with Trigonometric Functions Study the function f(x)=cosx+ 1 2x on[0,π] and sketch its graph; We have f (x)= sinx+ 1 2 ; Therefore, f(x) has critical points at x= π 6, 5π 6 ; We have f (x)= cosx; Therefore, f (x) has zero x= π 2 ; The following is a combined sign table for f (x),f (x) and summarizes the monotonicity and concavity of f(x): Function 0<x< π π 6 6 <x<π π 2 2 <x<5π 5π 6 6 <x<π f (x) + + f (x) + + f(x) It shows that f(0),f( 5π 6 ) are local minima, f( π 6 ),f(π) are local maxima and at x=π 2 f has inflection points; George Voutsadakis (LSSU) Calculus I November / 77

52 Graph Sketching and Asymptotes Graph Sketching with Exponential Functions Study the function f(x)=xe x and sketch its graph; We have f (x)=e x +xe x =(1+x)e x ; Therefore, f(x) has critical points at x= 1; We have f (x)=e x +e x +xe x =(2+x)e x ; Therefore, f (x) has zero x= 2; The following is a combined sign table for f (x),f (x) and summarizes the monotonicity and concavity of f(x): Function x< 2 2<x< 1 1<x f (x) + f (x) + + f(x) It shows that f( 1) are local minima and at x = 2 f has inflection point; Since lim x f(x)=0, x = 0 is a horizontal asymptote as x ; George Voutsadakis (LSSU) Calculus I November / 77

53 Graph Sketching and Asymptotes Graph Sketching with Rational Functions Study the function f(x)= 3x+2 2x 4 and sketch its graph; We have Dom(f)=R {2}; Since lim x 2 f(x)= and lim x 2 +f(x)=, the line x=2 is a vertical asymptote to y=f(x); Since lim x f(x)= 3 2 and lim x f(x)= 3 2, the line y=3 2 is a horizontal asymptote to y=f(x); We have f (x)= 3(2x 4) 2(3x+2) 4 = ; Therefore, f(x) has critical (2x 4) 2 (x 2) 2 point at x=2; We have f 8 (x)= ; Therefore, f (x) is undefined at (x 2) 3 x=2; The following is a combined sign table for f (x),f (x) and summarizes the monotonicity and concavity of f(x): George Voutsadakis (LSSU) Calculus I November / 77

54 Graph Sketching and Asymptotes Study of f(x)= 3x+2 2x 4 (Cont d) Recall f 4 (x)= (x 2) 2 and f (x)= 8 (x 2) 3. Function x<2 2<x f (x) f (x) + f(x) x= 2 is outside the domain of f; so it cannot be either a local extremum or an inflection point; Taking into account the arrows of the table and the asymptotes, we sketch the graph: George Voutsadakis (LSSU) Calculus I November / 77

55 Graph Sketching and Asymptotes Graphs of Rational Functions II Study the function f(x)= 1 x 2 1 and sketch its graph; We have Dom(f)=R { 1,1}; Since lim x 1 f(x)=, lim x 1 +f(x)= and lim x 1 f(x)=, lim x 1 +f(x)=, the lines x= 1 and x=1 are a vertical asymptotes to y=f(x); Since lim x f(x)=0 and lim x f(x)=0, the line y=0 is a horizontal asymptote to y=f(x); We have f (x)=[(x 2 1) 1 ] 2x = ; Therefore, f(x) has critical (x 2 1) 2 points at x=0,±1; We also have f (x) = 2(x2 1) 2 2x2(x 2 1)2x = 2(x2 1) 8x 2 (x 2 1) 4 (x 2 1) 3 = 6x2 2 = 2(3x2 +1) ; (x 2 1) 3 (x 2 1) 3 Therefore, f (x) is undefined at x=±1; The following is a combined sign table for f (x),f (x) and summarizes the monotonicity and concavity of f(x): George Voutsadakis (LSSU) Calculus I November / 77

56 Graph Sketching and Asymptotes Study of f(x)= 1 x 2 1 (Cont d) Recall f (x)= 2x (x 2 1) 2 and f (x)= 2(3x2 +1) (x 2 1) 3. Function x< 1 1<x<0 0<x<1 1<x f (x) + + f (x) + + f(x) x=±1 are outside the domain of f; so they cannot be either local extrema or inflection points; Taking into account the arrows of the table and the asymptotes, we sketch the graph: George Voutsadakis (LSSU) Calculus I November / 77

57 Applied Optimization Subsection 7 Applied Optimization George Voutsadakis (LSSU) Calculus I November / 77

58 Applied Optimization Rectangle of Fixed Perimeter with Maximum Area A piece of wire of length L is bent into a rectangular shape. Which dimensions produce a rectangle with maximum possible area? The objective function to be maximized is A = lw; The auxiliary condition 2l+2w=L allows us to reduce w by solving for it: w= 1 2L l; Thus A(l)=l( 1 2 L l)= l Ll; Compute derivative A (l)= 2l+ 1 2L; Find critical point: 2l+ 1 2 L=0 implies l= 1 4L; Thus, the rectangle must be of dimensions 1 4 L 1 4 L and it will have max area A max= 1 16 L2 ; George Voutsadakis (LSSU) Calculus I November / 77

59 Applied Optimization Minimizing Travel Time George is swimming a quarter mile from shore and sees Jessica on the beach two miles down from his closest point to the beach. He must reach her as soon as possible in case she leaves. If he can run a mile in 10 minutes and swim a mile in 40 minutes how can he get to Jessica in the least time possible? Suppose that George swims to a point x miles down the beach from the closest point and then runs the remaining distance on the beach. The objective function to be maximized is time T(x)= T (x)=40 40x=10 x x x 1 10 x =40 x (x 2); The derivative is 10= 40x 10 x x x ; so 16x2 =x ; Set that equal to zero and solve, i.e., x= miles; George Voutsadakis (LSSU) Calculus I November / 77

60 Applied Optimization Optimizing Price for Maximum Profit An apartment building has 30 units all of which are rented at $ 1,000 per month. For each $ 40 increase in rent an additional unit becomes vacant. Assume the maintenance cost per occupied unit is $ 120 per month. What is the rental price that maximizes the monthly profit? To devise an objective function for the profit, set x be the number of $ 40 increments and subtract revenue from cost: price/unit # of units maintenace costs P(x) = R(x) C(x)= ( x)(30 x) 120(30 x) = 40x x x = 40x x Thus, the maximum profit occurs when P (x)=0, i.e., when 80x+320=0, giving x=4. Thus, the most profitable rental price is =$1160 per month. George Voutsadakis (LSSU) Calculus I November / 77

61 Applied Optimization Can of Fixed Volume Using Least Aluminum What are the dimensions of a cylindrical can of volume 900 cm 3 that uses the least amount of aluminum, i.e., that has the minimum surface area? top & bottom The objective function is A= 2πr 2 + 2πrh; The volume is πr 2 h=900, whence, we get that h= 900 ; This allows us to express the objective πr 2 function as side A(r)=2πr ; r This has derivative A (r)=4πr 1800 r 2 r=0, π = 4πr r 2 and has critical points ; The second yields the min area; What is the required h? George Voutsadakis (LSSU) Calculus I November / 77

62 Newton s Method Subsection 8 Newton s Method George Voutsadakis (LSSU) Calculus I November / 77

63 Newton s Method Newton s Method I Suppose that f(x) is a continuous function and that we would like to find a root of the equation f(x)=0; This root is the x-intercept of the graph of y=f(x); Assume, in addition, that we have an initial guess x 0 close to the real solution; George Voutsadakis (LSSU) Calculus I November / 77

64 Newton s Method Newton s Method II The slope of the tangent line at x 0 is f (x 0 ); Thus, an equation of the tangent line at x 0 is y f(x 0 )=f (x 0 )(x x 0 ); Hence, its x-intercept x 1 is x 1 =x 0 f(x 0) f (x 0 ) ; The slope of the tangent line at x 1 is f (x 1 ); Thus, an equation of the tangent line at x 1 is y f(x 1 )=f (x 1 )(x x 1 ); Hence, its x-intercept x 2 is x 2 =x 1 f(x 1) f (x 1 ) ; We iterate this process, thus obtaining Newton s formula for approximating the roots of f(x)=0; George Voutsadakis (LSSU) Calculus I November / 77

65 Newton s Method Newton s Approximation Formula Newton s Method To compute a root of f(x)=0: Choose an initial guess x 0 suspected to be close to root; Generate successive approximations using x n+1 =x n f(x n) f (x n ) ; George Voutsadakis (LSSU) Calculus I November / 77

66 Newton s Method Approximating the 3 To approximate the 3, we consider the function f(x)=x 2 3 and try to find its zero; We have f (x)=2x; Let us take x 0 =2; Then we obtain successively: x 1 x 2 x 3 = 2 f(2) f (2) =2 1 4 =1.75 = 1.75 f(1.75) f (1.75) = =1.732 = f(1.732) f (1.732) = = Note that, using a calculator, we get ; George Voutsadakis (LSSU) Calculus I November / 77

67 Newton s Method Approximating a root of cos3x=sinx Consider the function f(x)=sinx cos3x and try to find a zero; We have f (x)=cosx+3sinx3x; Let us take x 0 =1; Then we obtain successively: x 1 x 2 x 3 x 4 = 1 f(1) f (1) =1 = 0.9 f( 0.9) f ( 0.9) = = = = f( 0.717) f ( 0.717) = = = f( 0.779) f ( 0.779) = = George Voutsadakis (LSSU) Calculus I November / 77

68 Antiderivatives Subsection 9 Antiderivatives George Voutsadakis (LSSU) Calculus I November / 77

69 Antiderivatives Antiderivatives A function F(x) is an antiderivative of f(x) on the interval(a,b) if F (x)=f(x) for all x (a,b); What is an antiderivative of f(x)=cosx? F(x)=sinx; Is it the only one? What is an antiderivative of f(x)=x 2? F(x)= 1 3 x3 ; Is it the only one? The General Antiderivative If F(x) is an antiderivative of f(x) on(a,b), then every other anti-derivative on(a,b) is of the form F(x)+C for some constant C. What are the most general antiderivatives of f(x)=x 6 and g(x)=sinx? F(x)= 1 7 x7 +C; G(x)= cosx+c; George Voutsadakis (LSSU) Calculus I November / 77

70 Antiderivatives Integration and Indefinite Integrals The process of determining an antiderivative is called integration; Indefinite Integral If F (x)=f(x), we write f(x)dx=f(x)+c; The most general antiderivative F(x) + C is called the indefinite integral of f(x); Find the indefinite integral x n dx, for n 1. x n dx= 1 n+1 xn+1 +C; George Voutsadakis (LSSU) Calculus I November / 77

71 Antiderivatives Power Rule for Indefinite Integrals Power Rule x n dx= 1 n+1 xn+1 +C, if n 1; Example: Use the power rule to compute the indefinite integrals: x 12 dx= 1 13 x13 +C 1 x 7 dx= x 7 dx= 1 6 x 6 +C= 1 6x 6 +C x 4/7 dx= 1 11/7 x11/7 +C= 7 11 x11/7 +C 1 5 x 3 dx= x 3/5 dx= 1 2/5 x2/5 +C= x2 +C George Voutsadakis (LSSU) Calculus I November / 77

72 Antiderivatives y= 1 x, Sum and Constant Factor Rules Antiderivative of y= 1 x 1 x dx=ln x +C; Linearity of the Indefinite Integral Sum/Difference Rule: (f(x)±g(x))dx= f(x)dx± g(x)dx Constant Factor Rule: cf(x)dx=c f(x)dx Example: Use the rules to evaluate the indefinite integral: (5x 2 3x 4/5 +x 9 )dx= 5x 2 dx 3x 4/5 dx+ x 9 dx= 5 x 2 dx 3 x 4/5 dx+ x 9 dx=5 1 3 x x9/5 1 8 x 8 +C= 5 3 x3 5 3 x9/5 1 8 x 8 +C; George Voutsadakis (LSSU) Calculus I November / 77

73 Antiderivatives Basic Trigonometric Integrals The following formulas give the basic trigonometric integrals: sinxdx= cosx+c sec 2 xdx= tanx+c cosxdx= sinx+c csc 2 xdx= cotx+c secxtanxdx= secx+c cscxcotxdx= cscx+c Based on the basic formulas, we obtain: cos(kx+b)dx= 1 k sin(kx+b)+c sin(kx+b)dx= 1 k cos(kx+b)+c Example: Evaluate: (sin(8x 3)+20cos9x)dx= sin(8x 3)dx+20 cos9xdx= 1 8 cos(8x 3) sin9x+c; George Voutsadakis (LSSU) Calculus I November / 77

74 Antiderivatives Integrals Involving e x Integrals Involving e x e x dx=e x +C e kx+b dx= 1 k ekx+b +C Example: (7e x 4x 3 )dx=7 e x dx 4 x 3 dx=7e x 4 x4 4 +C=7ex x 4 +C 12e 7 3x dx=12 e 7 3x dx= e7 3x +C= 4e 7 3x +C George Voutsadakis (LSSU) Calculus I November / 77

75 Antiderivatives Differential Equations and Initial Conditions An antiderivative of a function f(x) is a solution to the differential equation dy dx =f(x); The most general antiderivative of the form f(x)dx=f(x)+c involving C is called the general solution of the equation; By imposing an initial condition, i.e., some equation of the form y(x 0 )=y 0, for some specific values x 0 and y 0, one may specify a particular solution of the differential equation, i.e., enable the specification of a value for C; dy The group{ dx =f(x) (differential equation plus initial condition) y(x 0 )=y 0 is called an initial value problem; George Voutsadakis (LSSU) Calculus I November / 77

76 Antiderivatives Solving an Initial Value Problem Solve dy dx =4x7 subject to the initial condition y(0)=4; We have y(x)= 4x 7 dx=4 x 7 dx=4 x8 8 +C=1 2 x8 +C; Since y(0)=4, we get 4=0+C, i.e., C=4; Therefore, we obtain the particular solution y(x)= 1 2 x8 +4; Solve dy dt =sin(πt) subject to the initial condition y(2)=2; We have y(t)= sin(πt)dt= 1 π cos(πt)+c; Since y(2)=2, we get 2= 1 π cos(2π)+c, i.e., C=2+ 1 π ; Therefore, we obtain the particular solution y(t)= 1 π cos(πt)+2+ 1 π. George Voutsadakis (LSSU) Calculus I November / 77

77 Antiderivatives Applied Initial Value Problem A car traveling with velocity 50 m/s begins to slow down at t=0 with a constant deceleration of a= 10 m/s 2. What is the velocity v(t) at time t and what distance will be traveled before the car stops? We have v(t)= 10dt= 10t+C; Since v(0)=50, we get C=50, i.e., v(t)= 10t+50; Therefore, setting v=0, we find that it will take t=5 s for the car to stop; Moreover, s(t)= v(t)dt= ( 10t+50)dt= 5t 2 +50t+C; Since s(0)=0, we get C=0, i.e., s(t)= 5t 2 +50t; Therefore, setting t=5, we find that the car will travel s(5)=125 m before it comes to a halt; George Voutsadakis (LSSU) Calculus I November / 77